c
In loop $(1)$
$4-0.8 I-0.8\left(I-I_{1}\right)=0$
$2 I-I_{1}=5 \ldots(1)$
In loop $(2)$
$4-0.8 I_{1}+0.8\left(I-I_{1}\right)=0$
$-I+2 I_{1}=5 \ldots(2)$
From $(1)$ and $(2)$ $I_{1}=I=5 \mathrm{Amp}$
Voltage drop across any battery
$V=E-I r=1-5 \times 0.2=0 V$
